■数直線上の集合(その63)

 高貴な数とは、連分数v=[a0:a1,a2,・・,an,1,1,1,・・・]によって定義される。

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 [a0:a1,a2,・・,an]のk番目[a0:a1,a2,・・,ak]で打ち切ったとき、

近似分数の分子と分母をそれぞれ、Ak, Bkで表すと

  v=(τAn+An-1)/(τBn+Bn-1)

のように表現できる。

(81τ+334)/(73τ+301)=1+(8τ+33)/(73τ+301)

(8τ+33)/(73τ+301)=(16τ+66)/(146τ+602)

=(8√5+74)/(73√5+675)

=-5156/(73√5+675)(8√5-74)

=5156/(47030+2√5)

=2578/(23515+√5)

=1/(9+(313+√5)/2578)

=1/(9+97964/2578(313-√5))

=1/(9+97964/806914-2578√5))

=1/(9+1/(8+(23202-2578√5)/97964)

=1/(9+1/(8+(11601-1289√5)/48982)

=1/(9+1/(8+(126275596/48982(11601+1289√5))

=1/(9+1/(8+(126275596)/(568240182+63137798√5)

=1/(9+1/(8+1/(4+(63137798+63137798√5)/(126275596)

=1/(9+1/(8+1/(4+(1+√5)/2)

=1/(9+1/(8+1/(5+(√5-1)/2)

=1/(9+1/(8+1/(5+1/(√5+1)/2)

=1/(9+1/(8+1/(5+1/(1+(√5-1)/2)

=1/(9+1/(8+1/(5+1/(1+1/(√5+1)/2)

1だけずれているが・・・

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33・73-8・301=1

9・8-1・73=-1

33/301=1/(9+4/33)

33/301=1/(9+1/(8+1/4)

したがって、[0:a1a2,・・,an]→[0:a1a2,・・,an+1,1,1,1,・・・]

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